P6825 「EZEC-4」求和

发布于 2021-02-01 | 分类于莫比乌斯反演 | 3分钟 | 466字数 | 浏览量：

\sum_{i=1}^n \sum_{j=1}^n \gcd(i,j)^{i+j}

\sum_{d=1}^n\sum_{i=1}^n\sum_{j=1}^n [\gcd(i,j)=d]d^{i+j}

\sum_{d=1}^n\sum_{i=1}^{\lfloor \frac{n}{d} \rfloor }\sum_{j=1}^{\lfloor \frac{n}{d} \rfloor } [\gcd(i,j)=1]d^{(i+j)d}

\sum_{d=1}^n \sum_{t=1}^{\lfloor \frac{n}{d} \rfloor} \mu(t) \sum_{i=1}^{\lfloor \frac{n}{dt} \rfloor }\sum_{j=1}^{\lfloor \frac{n}{dt} \rfloor } d^{(i+j)dt}

\sum_{d=1}^n \sum_{t=1}^{\lfloor \frac{n}{d} \rfloor} \mu(t) (\sum_{i=1}^{\lfloor \frac{n}{dt} \rfloor } d^{idt})^2

\sum_{d=1}^n d^d \sum_{t=1}^{\lfloor \frac{n}{d} \rfloor} \mu(t) d^t (\sum_{i=1}^{\lfloor \frac{n}{dt} \rfloor } d^i)^2

\sum_{d=1}^n d^d \sum_{t=1}^{\lfloor \frac{n}{d} \rfloor} \mu(t) d^t (\frac{1-d^{\lfloor \frac{n}{dt} \rfloor}}{1-d})^2

\sum_{T=1}^n \sum_{d|T} d^T \mu(\frac{T}{d}) (\sum_{i=1}^{\lfloor \frac{n}{T} \rfloor }d^{i})^2

\sum_{T=1}^n \left( \sum_{d|T} d^T \mu(\frac{T}{d})(\frac{1-d^{\lfloor \frac{T}{d} \rfloor}}{1-d})^2 \right)

感谢您的支持，我会继续努力的!